# An electromegnet can lift a load of 45.0kg when it operates with a current of 5.0A. what current must the magnet be if it is to lift a 150kg load?

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### 1 Answer

This question comes down to two proportionalities given ideal conditions:

1) The strength of a magnetic field is proportional to the current in the electromagnet's coils.

2) The force exerted by a magnetic field is proportional to the square of the strength of the magnetic field.

Before we keep going, I'm just going to list a few variables:

B = magnetic field

I = current in wire

a,b,c = proportionality constants

F = force

Based on our first proportional relationship, we can say:

B = aI

Based on the second, we get:

F = bB^2

We'll just substitute for "aI" to get our force-current relationship:

F = b(aI)^2 = ba^2I^2

Now, ba^2 is a combination of two arbitrary (random) constants, so we'll just combine those into another arbitrary (random) constant, "c":

F = cI^2

Now, we can figure out what our arbitrary constant is based on the first conditions given in the problem. Keep in mind, to get F, we'll need the weight of the object, which will be "mg" where gravitational acceleration is 9.8 m/s^2. Our equation with F = mg becomes:

F = cI^2

45.0*9.8 = c(5.0)^2

441 = 25c

17.64 = c

Now that we have are not-so-arbitrary-anymore constant, we can figure out the current in the second situation using the same equation (Again, F = mg):

F = cI^2

150 * 9.8 = 17.64 I^2

1470 = 17.64 I^2

83.333 = I^2

9.1 = I

There we have it: our current in the magnet must be at least **9.1 amps.**

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